cos^2(4x)-sin^2(4x)=cos(b)

Solution

cos^{2} (4 x) - sin^{2} (4 x) = cos (b)

x = \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}, x = - \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}

Solution steps

cos^{2} (4 x) - sin^{2} (4 x) = cos (b)

Rewrite using trig identities

cos (2 \cdot 4 x) = cos (b)

Apply trig inverse properties

2 \cdot 4 x = arccos (cos (b)) + 2 πn, 2 \cdot 4 x = - arccos (cos (b)) + 2 πn

Solve

2 \cdot 4 x = arccos (cos (b)) + 2 πn : x = \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}

Solve

2 \cdot 4 x = - arccos (cos (b)) + 2 πn : x = - \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}

x = \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}, x = - \frac{arccos ( cos ( b ) )}{8} + \frac{πn}{4}

y, sin (x + y) + sin (y + z) + sin (x + z) = 0

2 sin (x) - tan (x) = 0

2 sin^{2} (x) - 13 sin (x) + 6 = 0

2 tan (θ) + 3 = 0

tan (x) = \frac{4}{6}